# What should be the distance between the object in Exercise 9.30

Question:

What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Solution:

Area of the virtual image of each square, A = 6.25 mm2

Area of each square, A0 = 1 mm2

Hence, the linear magnification of the object can be calculated as:

$m=\sqrt{\frac{A}{A_{0}}}$

$=\sqrt{\frac{6.25}{1}}=2.5$

But $m=\frac{\text { Image distance }(v)}{\text { Object distance }(u)}$

$\therefore v=m u$

$=2.5 u$    ...(1)

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{2.5 u}-\frac{1}{u}=\frac{1}{u}\left(\frac{1}{2.5}-\frac{1}{1}\right)=\frac{1}{u}\left(\frac{1-2.5}{2.5}\right)$

$\therefore u=-\frac{1.5 \times 10}{2.5}=-6 \mathrm{~cm}$

And $v=2.5 u$

$=2.5 \times 6=-15 \mathrm{~cm}$

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.